Given two positive integers left
and right
, find the two integers num1
and num2
such that:
left <= nums1 < nums2 <= right
.nums1
andnums2
are both prime numbers.nums2 - nums1
is the minimum amongst all other pairs satisfying the above conditions.
Return the positive integer array ans = [nums1, nums2]
. If there are multiple pairs satisfying these conditions, return the one with the minimum nums1
value or [-1, -1]
if such numbers do not exist.
A number greater than 1
is called prime if it is only divisible by 1
and itself.
Example 1:
Input: left = 10, right = 19 Output: [11,13] Explanation: The prime numbers between 10 and 19 are 11, 13, 17, and 19. The closest gap between any pair is 2, which can be achieved by [11,13] or [17,19]. Since 11 is smaller than 17, we return the first pair.
Example 2:
Input: left = 4, right = 6 Output: [-1,-1] Explanation: There exists only one prime number in the given range, so the conditions cannot be satisfied.
Constraints:
1 <= left <= right <= 106
Companies: TikTok
Related Topics:
Math, Number Theory
Similar Questions:
// OJ: https://leetcode.com/problems/closest-prime-numbers-in-range
// Author: github.com/lzl124631x
// Time: O(NloglogN)
// Space: O(N)
class Solution {
public:
vector<int> closestPrimes(int left, int right) {
vector<bool> isPrime(right + 1, true);
vector<int> primes;
int bound = sqrt(right), minDiff = INT_MAX, first = -1;
for (int i = 2; i <= right; ++i) {
if (!isPrime[i]) continue;
if (i >= left) primes.push_back(i);
if (i > bound) continue;
for (int j = i * i; j <= right; j += i) isPrime[j] = false;
}
for (int i = 1; i < primes.size(); ++i) {
int d = primes[i] - primes[i - 1];
if (d < minDiff) {
minDiff = d;
first = i - 1;
}
}
return first == -1 ? vector<int>{-1,-1} : vector<int>{primes[first], primes[first + 1]};
}
};